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no enthalpy change across the momentum source |
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December 16, 2005, 05:47 |
no enthalpy change across the momentum source
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#1 |
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I found that no enthalpy change across the momentum source, should it be that way?
My domain is the constant area tube. I create the disk with disk diameter the same as tube diameter inside the tube. The disk and the tube are coaxial. There is the energy source inside the tube, but very far from the disk. So I use 'Thermal energy' option in 'Domain'. I assign the disk as the momentum sink (momentum source with negative sign). I use the general momentum source which specified in the axial direction only. Because there is no other external force around there, and the area of the tube is not change, so this momentum source should make the pressure change predominantly across the disk. I found that the pressure divided by disk thickness is the same as the value of momentum source, this seems correct to me. However the result file shows that no enthalpy change across the disk. This is surprised to me. (The enthalpy value used are the output from CFX Post) Do anyone know why the pressure change but the enthalpy doesn't change??? Thank you very much for your consideration. Atit |
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December 18, 2005, 17:27 |
Re: no enthalpy change across the momentum source
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#2 |
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Try running the total energy equation.
Ther thermal energy equation is an equation for internal energy equation which does not accout for this effect. |
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December 19, 2005, 03:33 |
Re: no enthalpy change across the momentum source
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#3 |
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Dear Neale,
I have tried to use 'Total Energy' option instead of 'Thermal Energy', but the solutions are the same. Now I think that to determine if the energy source is the heat or the work in energy equation of CFX is that if only the energy source assigned only, the energy is the heat and if there is energy and momentum source simultaneousely, that energy source is the work. Do you think so? By the way, I still wonder about my case. There is no change in enthalpy, kinetic energy, and potential energy. So what energy compensate the pressure drop of the flow? Regards, Atit |
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